# Creating a topology on a metrizable space.

If I have a topological space $(E,\tau)$ that is metrizable for a metric $d$, and then I look at the set of all open balls in this metric $d$, will this always form a topology? If yes, will this topology be the same $\tau$ ?

• Joseph's answer differ (edit not anymore) – user286485 Jan 14 '17 at 16:33
• In general, the collection of open balls will not be a topology. Let's think to the case of the standard topology on $\mathbb{R}$ (which is induced by the metric $(x,y)\mapsto\vert x-y\vert$) ... However, the unit balls form a base for the topology induced by a metric (by definition !), which means that every open set is the union of a family of open balls. – Adren Jan 14 '17 at 16:35

• Yes, the topology induced by the open balls for the metric $d$ will be a topology.
• But no, there is no reason that this topology would be the same as $\tau$.
Take for instance $E=\mathbb R$, $d=\vert .\vert$ and $\tau=\mathfrak P(\mathbb R)$ the discrete topology where every subset is open.